# Find the Product

• Apr 20th 2009, 04:38 AM
magentarita
Find the Product
I need to find the product:

(1-1/2)(1-1/3)(1-1/4)(1-1/5)......(1-1/499)(1-1/500) =

NOTICE: The list goes from (1-1/2) all the way to
(1 - 1/500).

How do I solve this question?

How do I find the product?
• Apr 20th 2009, 04:49 AM
Quote:

Originally Posted by magentarita
I need to find the product:

(1-1/2)(1-1/3)(1-1/4)(1-1/5)......(1-1/499)(1-1/500) =

NOTICE: The list goes from (1-1/2) all the way to
(1 - 1/500).

How do I solve this question?

How do I find the product?

1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500
• Apr 20th 2009, 05:39 AM
e^(i*pi)
Quote:

1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500

Indeed you do. For n terms this will be

$\frac{(n-1)!}{n!} = \frac{1}{n}$
• Apr 20th 2009, 09:01 PM
magentarita
tell me
Quote:

1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500

How do I use the formula (1 - n)!/n! = 1/n to find the same answer?
• Apr 20th 2009, 09:02 PM
magentarita
tell me...
Quote:

Originally Posted by e^(i*pi)
Indeed you do. For n terms this will be

$\frac{(n-1)!}{n!} = \frac{1}{n}$

How do I use the formula (1 - n)!/n! = 1/n to find the same answer?
• Apr 20th 2009, 10:03 PM
Quote:

Originally Posted by magentarita
How do I use the formula (1 - n)!/n! = 1/n to find the same answer?

http://www.mathhelpforum.com/math-he...7000fa16-1.gif

$=\frac{(500-1)(500-2).....(2)(1)}{(500)(499)...(2)(1)}$

$=\frac{(500-1)!}{500!}$

$=\frac{(499)!}{500!}$

Remember that (n-1)! is not equal to (1-n)!

$=\frac{(n-1)!}{n!} = \frac{1}{n} {\color{red}\ne} \frac{(1-n)!}{n!}$

$=\frac{1}{500}$
• Apr 21st 2009, 09:01 PM
magentarita
I see...
Quote:

http://www.mathhelpforum.com/math-he...7000fa16-1.gif

$=\frac{(500-1)(500-2).....(2)(1)}{(500)(499)...(2)(1)}$

$=\frac{(500-1)!}{500!}$

$=\frac{(499)!}{500!}$

Remember that (n-1)! is not equal to (1-n)!

$=\frac{(n-1)!}{n!} = \frac{1}{n} {\color{red}\ne} \frac{(1-n)!}{n!}$

$=\frac{1}{500}$

I see what I did wrong.